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Polymerization equilibrium. Consider the process of polymerization and assume the principle of equal reactivity, that each monomer adds with the same equilibrium constant as the previous one: 2X1KX2,3X1K2X3,nX1Km1Xn. (a) Write the binding polynomial Q. (b) Write the average chain length n in terms of Q. (c) Plot n versus x for K=1 and for Kx<1.

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01

- Write the binding polynomial Q

The binding polynomial (also known as the partition function) is given by the sum of all possible states that the system can adopt. For the polymerization process, each state corresponds to the formation of a polymer chain of length n with an equilibrium constant that depends on n. Given the equilibrium constants, the binding polynomial can be written as: Q=sum of all states=1+K[X1]22!+K2[X1]33!+K3[X1]44!+K4[X1]55!+ up to n terms
02

- Write the average chain length ⟨n⟩ in terms of Q

The average chain length n is the average number of monomers in the polymer chain. It can be expressed in terms of the binding polynomial Q as: n=total sum of chain lengthsQ =n=1and termn×QnQ

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

binding polynomial
In polymerization equilibrium, the binding polynomial, also known as the partition function, is an essential concept. It represents the sum of all possible states the polymer system can adopt.
For polymer chains, each state corresponds to a polymer of a certain length forming with a specific equilibrium constant. Given the equilibrium constants, the binding polynomial can be formulated as follows:

The basic form of the binding polynomial is:
Q=1+K[X1]22!+K2[X1]33!+K3[X1]44!+K4[X1]55!+ up to n terms

Here, the first term 1 represents the monomer itself, and each successive term represents polymer chains of length increasing by one monomer at a time.

This polynomial helps us understand how the polymers of various lengths contribute to the overall equilibrium.
Remember that in these expressions, K is the equilibrium constant for the polymerization step, and [X1] is the concentration of the monomer.
This comprehensive sum describes all possible configurations from monomers to large chains.
average chain length
The average chain length n signifies the mean number of monomers in a polymer chain. This is crucial as it indicates the extent of polymerization in the system.
To determine this, we consider the binding polynomial Q and the sum of all chain lengths weighted by their occurrences:
n=total sum of chain lengthsQ=n=1end termn×QnQ
This formula breaks down as follows:
  • n represents the number of monomers in the chain.
  • Qn is the term representing the weight of a polymer chain of length n.
  • Q is the binding polynomial, which normalizes the total sum.
For instance, if you have chains up to length 5, the terms within the summation would be 1×Q1,2×Q2, and so forth. This calculation ultimately gives the expected length of polymers in the system.
Thus, seeing changes in n under different conditions can provide insights into how the polymerization process behaves.
equilibrium constants
Equilibrium constants (K) play a central role in understanding polymerization equilibrium.
Each step in the polymerization process has an associated equilibrium constant, which describes the ratio of the concentrations of products to reactants at equilibrium. For polymer chains:
  • 2X1KX2
  • 3X1K2X3
  • ... and so forth.
In this scheme, the equilibrium constant for the formation of a dimer is K, a trimer involves K2, and so on.
These constants highlight the progressive nature of polymerization, where each additional monomer joins with a dependence on the previous equilibrium constants.
Understanding these equilibrium constants is vital as they affect the formation and stability of different polymer chain lengths, directly influencing the binding polynomial (Q) and the average chain length n.

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Most popular questions from this chapter

Drug binding to a protein. A drug D binds to a protein P with two different equilibrium binding constants, K1 and K2 : D+PK1PD1, and D+PD1K2PD2, where K1=1000M1, and K2=2000M1 at T=300 K. (a) Write an algebraic expression for the fraction of protein molecules that have two drug molecules bound as a function of drug concentration x=[D]. (b) If the drug concentration is x=103M, what is the fraction of protein molecules that have one drug molecule bound? (c) If the drug concentration is x=103M, what is the fraction of protein molecules that have zero drug molecules bound? (d) If the drug concentration is x=103M, what is the fraction of protein molecules that have two drug molecules bound?

Scatchard plots don't work for multiple types of binding site. If a binding process involves n1 sites with affinity K1, a Scatchard plot of v/x versus v gives a straight line with K1 as slope and n1 as horizontal-axis intercept (see Figure 28.6). Show that if you have two types of sites having (number, affinity) =(n1,K1) and (n2,K2), a Scatchard plot doesn't give two straight lines from which you can get (n2,K2). To fit multiple types of sites, you need a different model.

Three-site binding. A ligand X can bind to a macromolecule P at three different binding sites with binding constants K1,K2, and K3 : X+PK1PX,X+PXK2PX2, and X+PX2K3PX3. (a) Write the binding polynomial, Q. (b) Write an expression for the number of ligands v bound per P molecule. (c) Compute v for x=[X]=0.05, assuming K1=1, K2=1, and K3=1000. (d) Assume the same K values as in (c). Below ligand concentration x=x0, most of the macromolecular P molecules have zero ligands bound. Above x=x0, most of the P molecules have three ligands bound. Compute x0. (e) For x=x0 in (d), show the relative populations of the ligation states with zero, one, two, and three ligands bound.

Relating stoichiometric to site constants. Derive the relationship between stoichiometric-model binding constants K1 and K2 and site-model binding constants Ka and Kb. Suppose the sites are independent, Kc=KaKb.

A model for Alzheimer's fibrillization. Consider a solution of protein molecules in three possible states of aggregation: in state A1, the protein is a monomer in solution; in state A10, the protein is hydrophobically clustered into oligomers having 10 molecules each; and in state A100,100 protein molecules are all aggregated into long fibrils. The equilibrium is given by 10A1Kolim A10100A1Kmpel A100 (a) Express the binding polynomial Q for this association equilibrium in terms of Kallgo, Ktharil , and x, where x is the concentration of protein A. (b) Write an expression for the average number v of protein molecules per object (monomer, oligomer, or fibril).

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